1. Continuous Function

for which value of the constant $K$ is the function $h(x)=\left\{\begin{array}{cc}\frac{e^x -e^a}{x-a},&\mbox{ if }
x\ne a\\K, & \mbox{ if } x=a\end{array}\right.$

continuous at $x=a$. justify your answer.

$\lim_{x \rightarrow a} h(x) = \lim_{x \rightarrow a} \frac{e^x - e^a}{x-a}$

$h(a) = K$ and the other limit is $e^a \log(e)$

so is $K = e^a \log(e)$?, is this a correct way of answering the question

2. Originally Posted by iLikeMaths
for which value of the constant $K$ is the function $h(x)=\left\{\begin{array}{cc}\frac{e^x -e^a}{x-a},&\mbox{ if }
x\ne a\\K, & \mbox{ if } x=a\end{array}\right.$

continuous at $x=a$. justify your answer.

$\lim_{x \rightarrow a} h(x) = \lim_{x \rightarrow a} \frac{e^x - e^a}{x-a}$
$h(a) = K$ and the other limit is $e^a \log(e)$
so is $K = e^a \log(e)$?, is this a correct way of answering the question
yes, but log should be natural log, and also $\ln(e)=1$.
3. so $K =e^a$ thanks